Chemistry

[xtullyx16]

Need help with a chemistry problem guys!

Here it is:

An olympic sized pool is 50.0m long and 25.0 m wide.

a) how many gallons of water (density of water = 1/0 g/mL) are needed to fill the pool to an average depth of 5.5 ft

My math:

5.5ft --> m

5.5 ( 12 in/1 ft)(2.54 cm/1 in)( 10 ^ -2m/ 1 cm) = 1.6764 m

Volume of Pool = L x W x H

= 50.0 x 25.0 m x1.6764 m
= 2095.5 m ^3 now i want this value in gallons...

2095.5 m^ 3 ( 1L/ 10^ -3 m^3)( 1 gal/ 3.7854 L)
= 553,574 gal.
= 5.5 x10 ^5 gal

B. What is the mass in (kg) of water in the pool?

Density= Mass/Volume

Density=1.0 g/mL
Volume 5.5 x10 ^5 gallons

Gal--> mL

(5.5x10^5/1)(3.7854L/ 1 gal)(10^3 cm^3/ 1 L)( 1 mL/ 1 cm^3) = 2.08197 x 10^9 mL

density Grams/mL --> kg/mL

1.0g (10^-3 kg/ 1 g) = .001 kg/ml

D= M/V

.001 kg/mL = mass/ 2.08197 x 10^9 mL

Mass = .001kg/2.08197 x 10^9 mL

Mass = 2.1 x 10^ 6 kg

All conversions were found in back of my chem book.. just checking if it all makes sense!

[250ex_dan]

try mathway.com

[xtullyx16]

I know the calculations are correct if i used the dimensional analysis right.. but i was thinking some engineers on here could lend a helping hand. I just dont really know about the part b.. but id like to make sure my DA is correct on part A before i move on!

[beastlywarrior]

try this, 1mL=1 gram

[wilkin250r]

Originally posted by xtullyx16


B. What is the mass in (kg) of water in the pool?

Density= Mass/Volume

Density=1.0 g/mL
Volume 5.5 x10 ^5

confused from there on: A) mass should be in terms of kg but if i calculate that out ( converting volume into mL ) then ill have g/ml ^2 which then how do convert to kg's?

Ok, I'll be honest, I didn't double-check your math. You said it was correct, so I'll just believe you.

However, look at your equation. Density=Mass/Volume. Solve this equation for mass, you get Mass=Density*volume

(g/ml)*(ml)=g

You must have divided to get g/ml^2.

Do you really need me to double check the math?

[xtullyx16]

Wilkin, I actually took a second to think about it, and i think I have it correctly.. I wish i had an answer sheet so i could see if my answer is correct.. I'm just afraid my dimensional analysis is wrong... this stuff should be a cake walk but i havent used it in 3 years..:(

[wilkin250r]

Okay, I double-checked everything. So far, you're good on your dimensions.

The only problem I see is your significant digits, you've got some rounding errors in there.

You calculated volume at 5.5 x10^5, (two significant digits), but your conversion for Liters/Gallon is 3.7854, (5 digits) and you calculated your volume out to 2.08197x10^9 ml. (6 digits).

If you use your actual calculated volume of 553,574 instead of 5.5x10^5, your final calculation for volume would end up 2.09550x10^9

Now, 2.08197 vs 2.09550 isn't a real significant error (0.65%). However, what's the point of carrying out 2.08197 out to five decimal places, when your calculations are only accurate out to two?

It's like calculating the distance from San Fransisco to Boston. What's the point of calculating down to the millimeter, if your error is plus or minus a mile?

[xtullyx16]

Originally posted by wilkin250r
Okay, I double-checked everything. So far, you're good on your dimensions.

The only problem I see is your significant digits, you've got some rounding errors in there.

You calculated volume at 5.5 x10^5, (two significant digits), but your conversion for Liters/Gallon is 3.7854, (5 digits) and you calculated your volume out to 2.08197x10^9 ml. (6 digits).

If you use your actual calculated volume of 553,574 instead of 5.5x10^5, your final calculation for volume would end up 2.09550x10^9

Now, 2.08197 vs 2.09550 isn't a real significant error (0.65%). However, what's the point of carrying out 2.08197 out to five decimal places, when your calculations are only accurate out to two?

It's like calculating the distance from San Fransisco to Boston. What's the point of calculating down to the millimeter, if your error is plus or minus a mile?

Yeah my sig figs were off.. but I checked with my friend in class and we both shared the same answers when rounded to the correct significnt figures.. thanks for all the check work wilkin!

[Quad18star]

Way to steal my answers Eric (Wilkin) !!! :macho





:p

[wilkin250r]

Originally posted by Quad18star
Way to steal my answers Eric (Wilkin) !!! :macho

The answers aren't the same in Canadian, so I'm not sure you'd be much help anyways. When we need the chemistry for Jellied Moose-Nose and hockey rinks, we'll give you a call... :blah:

[Quad18star]

Originally posted by wilkin250r
The answers aren't the same in Canadian, so I'm not sure you'd be much help anyways. When we need the chemistry for Jellied Moose-Nose and hockey rinks, we'll give you a call... :blah:

LOL. I'll tell you exactly how much water you'll need for the rink. ;)